BZOJ1576: [Usaco2009 Jan]安全路经Travel(最短路 并查集)

题意

给你一张无向图，保证从1号点到每个点的最短路唯一。对于每个点求出删掉号点到它的最短路上的最后一条边(就是这条路径上与他自己相连的那条边)后1号点到它的最短路的长度

Sol

emmm，考场上想了个贪心开心的飞起然而只多得了10分qwq

正解比较神仙。

首先把最短路树建出来，考虑一条非树边$(u, v)$什么时候能更新答案

结论是：除了他们的LCA外的子树内其他都可以更新，且新的权值为$dis[u] + dis[v] + w(u, v) - dis[x]$，$x$表示新节点

这样我们把所有的边按照$dis[u] + dis[v] + w(u, v)$排序，显然，一个点如果被更新过那么就再也不会被更新了。

用并查集把已经更新过的点缩起来即可

这题的关键是要发现非树边与答案之间的性质。。

#include<cstdio>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
#define Pair pair<int, int> 
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M;
vector<Pair> v[MAXN];
struct Edge {
    int u, v, w;
}E[MAXN];
int num = 0;
int dis[MAXN], top[MAXN], vis[MAXN], cnt = 0, ans[MAXN];
void Dij() {
    memset(dis, 0x7f, sizeof(dis));
    dis[1] = 0;
    priority_queue<Pair> q; q.push(MP(0, 1));
    while(!q.empty()) {
        int p = q.top().se; q.pop();
        if(vis[p]) continue; vis[p] = 1;
        for(int i = 0; i < v[p].size(); i++) {
            int to = v[p][i].fi, w = v[p][i].se;
            if(dis[to] > dis[p] + w && (!vis[to])) {
                top[to] = p;
                dis[to] = dis[p] + w;
                q.push(MP(-dis[to], to));
            }
        }
    }
}
int comp(const Edge &a, const Edge &b) {
    return dis[a.u] + dis[a.v] + a.w < dis[b.u] + dis[b.v] + b.w;
}
int fa[MAXN];
int unionn(int x, int y) {
    fa[x] = y;
}
int Find(int x) {
    if(fa[x] == x) return fa[x];
    else return fa[x] = Find(fa[x]);
}
int solve(int x, int y, int w) {
    while((x = Find(x)) != (y = Find(y))) {
        //int dx = dis[x], dy = dis[y];
        if(dis[x] < dis[y]) swap(x, y);
        ans[x] = w - dis[x];
        x = (fa[x] = top[x]);
        cnt++;
    }
}
int main() {
    N = read(); M = read();    
    for(int i = 1; i <= M; i++) {
        int x = read(), y = read(), z = read();
        v[x].push_back(MP(y, z));
        v[y].push_back(MP(x, z));
    }
    Dij();
    for(int i = 1; i <= N; i++) {
        fa[i] = i;
        for(int j = 0; j < v[i].size(); j++) {
            int to = v[i][j].fi, w = v[i][j].se;
            if(top[to] == i || top[i] == to) continue;
            E[++num] = (Edge) {i, to, w};
        }
    }
    sort(E + 1, E + num + 1, comp);
    for(int i = 1; i <= num; i++) {
        solve(E[i].u, E[i].v, dis[E[i].u] + dis[E[i].v] + E[i].w);
        if(cnt == N - 1) break;
    }
    for(int i = 2; i <= N; i++)
        printf("%d\n", ans[i] ? ans[i] : -1);
    return 0;
}

标签：BZOJ1576,vis,int,Travel,查集,fa,MAXN,include,dis
来源： https://blog.51cto.com/u_15239936/2868696